Solution 1 to problem N1t6s11b4+5

Expressions | Parameters | Inequalities | Relevance | Back to problem N1t6s11b4+5

Expressions

The solution is given through the following expressions:


q28=0


q27=0


q26=0


q25=0


     p6*q9
q24=-------
      p1


q23=q9


q22=0


q21=0


q20=0


q19=0


q18=0


q17=q9


q16=0


q15=0


q14=0


q13=0


q12=0


q11=0


q10=0


q8=0


q7=0


q6=0


q5=0


q4=0


q3=0


q2=0


q1= - q9


p7=0


p5=p1


p4=0


p3= - p1


p2=0

Parameters

Apart from the condition that they must not vanish to give a non-trivial solution and a non-singular solution with non-vanishing denominators, the following parameters are free:

 q9, p1, p6

Inequalities

In the following not identically vanishing expressions are shown. Any auxiliary variables g00?? are used to express that at least one of their coefficients must not vanish, e.g. g0019*p4 + g0020*p3 means that either p4 or p3 or both are non-vanishing.

 
{p5,g0006*q9 - g0011*q9 - g0012*p1,q24,q9,p1,p6,g0036*p6 + g0037*p1 + g0043*p1,

 p3}

Relevance for the application:

The system:


          2
b(1) =b(2) *p6 + b(1) *b(1)*p1
    t                x

b(2) = - Db(2)*Db(1)*p1 + b(2) *b(1)*p1
    t                         x

The symmetry:

           3
       b(2) *p6*q9 + Db(2)*Db(1)*b(1)*p1*q9 + b(1) *b(2)*b(1)*p1*q9
                                                  x
b(1) =--------------------------------------------------------------
    s                               p1

b(2) = - Db(2)*Db(1)*b(2)*q9 + b(2) *b(2)*b(1)*q9
    s                              x

And now in machine readable form:

The system:

df(b(1),t)=b(2)**2*p6 + df(b(1),x)*b(1)*p1$

df(b(2),t)= - d(1,b(2))*d(1,b(1))*p1 + df(b(2),x)*b(1)*p1$

The symmetry:

df(b(1),s)=(b(2)**3*p6*q9 + d(1,b(2))*d(1,b(1))*b(1)*p1*q9 + df(b(1),x)*b(2)*b(1
)*p1*q9)/p1$

df(b(2),s)= - d(1,b(2))*d(1,b(1))*b(2)*q9 + df(b(2),x)*b(2)*b(1)*q9$